You have found the following ages (in years) of 5 tigers. Those tigers were randomly selected from the 47 tigers at your local zoo: $ 7,\enspace 5,\enspace 4,\enspace 17,\enspace 28$ Based on your sample, what is the average age of the tigers? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 47 tigers, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{7 + 5 + 4 + 17 + 28}{{5}} = {12.2\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {27.04} + {51.84} + {67.24} + {23.04} + {249.64}} {{5 - 1}} $ {s^2} = \dfrac{{418.8}}{{4}} = {104.7\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{104.7\text{ years}^2}} = {10.2\text{ years}} $ We can estimate that the average tiger at the zoo is 12.2 years old. There is also a standard deviation of 10.2 years.